Please help me with this. How do I prove that a square matrix A has non-zero determinant(nonsingular) if and only if the characteristic equation of A has all non-zero roots. Its probably very simple...
Thank you in advance
Please help me with this. How do I prove that a square matrix A has non-zero determinant(nonsingular) if and only if the characteristic equation of A has all non-zero roots. Its probably very simple...
Thank you in advance
The characteristic polynomial of $A$ is $p(\lambda)=\det (A-\lambda I)$. So, $p(0)=\det A$ and the result follows.